I'm learning Prolog and I'm having some difficulties to understand how this particular operator works. I've been told it's used to stop backtracking during recursivity, but I can't quite picture it. What would be an example in a non-declarative language like Java? Maybe if I find a parallel I'm used to I'll be able to understand it.
Thank you!
The cut (!/0) doesn't "stop backtracking": it simply eliminates a choice point. Consider something like this,
even_or_odd( X , even ) :- 0 =:= X mod 2 .
even_or_odd( _ , odd ) .
For any odd integer, it has a single solution: odd. But for any even integer, it has two solutions: it will first identify as even, and on backtracking, identify as odd.
An integer's being even or odd is deterministic: 4 is not going to be even on first evaluation, and then magically become odd on reevaluation.
So, we introduce a cut to eliminate the choice point(s)/alternative(s):
identify_even_or_odd( X , even ) :- 0 =:= X mod 2 , ! .
identify_even_or_odd( _ , odd ) .
Now, having made the determination that X is even, the cut eliminates the choice point, and essentially "prunes" the search tree: backtracking into the cut causes the entire containing predicate to fail (and continue backtracking through the solution tree).
